Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{-n^2 - 4n}{-3n^2 + 27n - 60} \div \dfrac{n^2 + 7n}{3n^2 + 6n - 105} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-n^2 - 4n}{-3n^2 + 27n - 60} \times \dfrac{3n^2 + 6n - 105}{n^2 + 7n} $ First factor out any common factors. $p = \dfrac{-n(n + 4)}{-3(n^2 - 9n + 20)} \times \dfrac{3(n^2 + 2n - 35)}{n(n + 7)} $ Then factor the quadratic expressions. $p = \dfrac {-n(n + 4)} {-3(n - 5)(n - 4)} \times \dfrac {3(n - 5)(n + 7)} {n(n + 7)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-n(n + 4) \times 3(n - 5)(n + 7) } { -3(n - 5)(n - 4) \times n(n + 7)} $ $p = \dfrac {-3n(n - 5)(n + 7)(n + 4)} {-3n(n - 5)(n - 4)(n + 7)} $ Notice that $(n - 5)$ and $(n + 7)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-3n\cancel{(n - 5)}(n + 7)(n + 4)} {-3n\cancel{(n - 5)}(n - 4)(n + 7)} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $p = \dfrac {-3n\cancel{(n - 5)}\cancel{(n + 7)}(n + 4)} {-3n\cancel{(n - 5)}(n - 4)\cancel{(n + 7)}} $ We are dividing by $n + 7$ , so $n + 7 \neq 0$ Therefore, $n \neq -7$ $p = \dfrac {-3n(n + 4)} {-3n(n - 4)} $ $ p = \dfrac{n + 4}{n - 4}; n \neq 5; n \neq -7 $